Calderón–Vaillancourt theorem

Remarks

One standard proof cuts the symbol into pieces $a_{jk}$ localized near the points $(j, k)$ of the lattice $\Z^n \times \Z^n$ by a smooth partition of unity. Each piece gives an operator with norm bounded by finitely many derivatives of $a$. Integration by parts shows that the products $T_{jk}^* T_{j'k'}$ and $T_{jk} T_{j'k'}^*$ decay fast in $\abs{j - j'}$ and $\abs{k - k'}$. The Cotlar–Stein lemma then sums the pieces.

Example: if $a(x, \xi) = m(\xi)$ does not depend on $x$, then $a(x, D)$ is the Fourier multiplier with symbol $m$, and Plancherel's theorem gives the bound $\sup \abs{m}$ directly. The theorem says that bounded smooth dependence on $x$ costs only finitely many derivatives. The same conclusion holds for symbols in the classes $S^0_{\rho, \rho}$ with $0 \le \rho < 1$.

For the class $S^0_{1,1}$, which allows $x$-derivatives to grow like $\abs{\xi}$, the operators are in general not bounded on $L^2$.

Going deeper

For a student

This is the basic $L^2$ boundedness result of the calculus of pseudo-differential operators. It shows that operators of order $0$ in the classes $S^0_{\rho,\delta}$ with $\delta \le \rho$ and $\delta < 1$ are bounded on $L^2$. By composition with $(I - \Delta)^{s/2}$, operators of order $m$ then map the Sobolev space $H^{s+m}$ to $H^s$. Its proof is a model use of almost orthogonality. The class $S^0_{0,0}$ is invariant under translations in both $x$ and $\xi$, which makes it natural in phase-space and semiclassical analysis.